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John Rumm John Rumm is offline
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Default Power factor, electricity meters and CFLs ?

Nick wrote:

Could someone explain power factor for me please (and I guess a few others)


It is a measure of how far out of phase the current being drawn from the
supply is with the voltage. With straight resistive loads the current
will be in phase with the voltage. The more capacitive or inductive
(i.e. the more reactive) the load appears, then the more out of phase
the current drawn will be (up to 90 degrees shift for a perfect inductor
or capacitor).

So a "unity" unity power factor of 1 (or 100%) is a resitive load, a
power factor of 0 is a fully reactive one.

The reason being is that I have a lot of CFLs and so the rated power
consumption is somewhat less than if these were ordinary filament bulbs.


This is typically true of all fluorescent lights unless corrected. They
normally have an inductive choke in series with the

However the other day I plugged in a few to my gizmo that tells power being
drawn in watts and also power factor.

The power factor was around 60% on the CFLs.

I believe the elec company doesn't like devices with a poor (low?) power
factor but I don't know why.


Imagine you are riding a bike up a hill. How hard you have to push on
the pedals will depend on a number of things - the steepness of the
hill, and how fast you want to get up it for example. Now say some evil
sod came along and modified your bike by attaching a large spring to the
right hand pedal, and fixing the other end to the seat post. Now every
time you push the right pedal down, you are not only overcoming the
resistance to move the bike, but you are stretching the spring. This
would make pushing the pedal harder. However when you go to push the
left pedal, the spring on the other side is now pulling and hence
helping. All the energy you stored in the spring on one cycle you now
get back on the other. The result is a bike that consumes the same
amount of real energy to do the same task, however it appears harder to
ride because the total peak push required is higher.

Poor power factors work the same way. The actual power dissipated will
remain unchanged, but the peak currents that flow around the circuit
will be higher.

Hence the generators need higher current capacity, and you suffer more
transmission losses in the distribution network. There is also a
additional problem that because reactive loads are drawing current that
is not in phase with the voltage waveform they can tend to distort the
waveform - taking lumps out of it. This has the effect of adding noise
to the mains supply for other users (which in turn can cause them some
power wastage since this harmonic noise is often dissipated as extra
heat in their appliances)

The other concern is how does the electricity meter measure power drawn by a
device with a low power factor ? - does it over or under read ? Also does
this vary between the older eddy current spinning discs and the newer
flashing LED meters with some other method of measuring power.


For domestic customers you are billed only for real power consumed - the
meter should be unaffected by non unity power factors. As to how good
the various models are in this respect I don't know.

Basically my elec bill is still way higher than I think it should be and I
am trying to get to the bottom of it.


It would seem unlikely that poor power factor is your problem. It does
become an issue for some industrial consumers since they are billed
based on the apparent power consumed.

--
Cheers,

John.

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